How much work would it take to horizontally accelerate an object with a mass of #8# #kg# to #4# #ms^-1# on a surface with a kinetic friction coefficient of #1 #?

2 Answers
Apr 16, 2016

Answer:

Not enough information provided.

Explanation:

We can divide the total work into two parts,
1. To increase the kinetic energy of the object. (Which is known with the formula #1/2 mv^2#)

  1. The work done against the frictional force. (Given as #mu*mg*s#; unknown as no other parameter except #v# (final velocity) is provided).
Apr 16, 2016

Answer:

#(64-156.8t)N#, where #t# is the time of acceleration.
With the condition that #t>0.4082s#

Explanation:

Force #F# applied to accelerate the object#=m.a#
Assuming that the object is initially at rest on the horizontal and acceleration is constant. Using the kinematic equation

#v=u+at#
#4=at#
#a=4/t#
Also distance moved #s=ut+1/2at^2#
#s=1/2xx4/txxt^2=2t#

Given is coefficient of kinetic friction #mu=1#.
As normally coefficient of static friction is greater than the coefficient of kinetic friction, also that neither of the coefficients can be greater than #1#, implies that both the coefficients #mu_s and mu_k=1#.

Let the normal reaction be #N=mg#, where gravity #g=9.8ms^-2#.
Force of friction which will oppose the motion #F_f=mu.mg#
Inserting given values
#F_f=1xx8xx9.8=78.4N#
#"Net Force"="Applied Force"-"Force due to friction"#
#"Net Force"=8xx4/t-78.4#
#=32/t-78.4#

Work done #="Force"xx"Distance"#

#=(32/t-78.4)xx2t#
#=64-156.8t#